Question

How do you graph `r = theta, theta in [ 0, 2pi ]`, using its Cartesian equivalent?

Answer 1

See explanation and graphs.

Explanation:

Range-restricted `abs r <= pi/2` combined graph, for `theta uarr`

in both clockwise and anticlockwise sense, using

`sqrt(x^2+y^2) = arctan(y/x)`

graph{sqrt(x^2+y^2) - arctan(y/x) = 0 }

Range-unrestricted combined graph, for `theta uarr`

in both clockwise and anticlockwise sense, using the wholesome

inverse

y = x tan (sqrt ( x^2 + y^2 ))

graph{(x tan ((x^2+y^2)^0.5)-y)(x-2pi+0.0001y)(x+2pi+0.0001y)=0}

It appears that the algorithm for the Cartesian

frame has to be improved, to separate the conjoined twin graphs.